A) n
B) \[\frac{n}{\sqrt{2}}\]
C) 2n
D) 4n
Correct Answer: A
Solution :
Key Idea: The frequency of vibrating wire is \[n=\frac{1}{2}\sqrt{\frac{T}{m}}\], where T is the tension in the wire, We have \[n=\frac{1}{2l}\sqrt{\frac{T}{m}}\] Here, m = mass per unit length \[=\pi {{r}^{2}}d\] \[\therefore n=\frac{1}{2\ell }\sqrt{\frac{T}{\pi {{r}^{2}}d}}\] \[orn\,\propto \,{{\left( \frac{T}{{{r}^{2}}d} \right)}^{1/2}}\] \[orn\,\propto \,\frac{1}{r}{{\left( \frac{T}{d} \right)}^{1/2}}\] \[\therefore \frac{{{n}_{1}}}{{{n}_{2}}}=\frac{{{r}_{2}}}{{{r}_{1}}}{{\left( \frac{{{T}_{1}}}{{{T}_{2}}}\times \frac{{{d}_{2}}}{{{d}_{1}}} \right)}^{1/2}}\] We have given, \[\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{1}{2},\,\frac{{{d}_{1}}}{{{d}_{2}}}=2,\,\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{1}{2}\] \[\therefore \frac{{{n}_{1}}}{{{n}_{2}}}=\frac{2}{1}{{\left( \frac{1}{2}\times \frac{1}{2} \right)}^{1/2}}\] \[or\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{2}{1}\times \frac{1}{2}=1\] \[or{{n}_{2}}={{n}_{1}}=n\]
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